Question: $\text E = \left[\begin{array}{rrr}1 & 4 & 2 \\ 1 & 2 & 2\end{array}\right]$ and $\text B = \left[\begin{array}{rr}1 & 3 \\ 2 & 3 \\ 2 & -2\end{array}\right]$ Let $\text {H = EB}$. Find $\text H$. $ {H = }$
Solution: The Strategy When multiplying matrices, we should find each entry of the resulting product matrix separately. To find entry $(i,j)$ of the resulting product matrix, we calculate the vector dot product of row $i$ of the first matrix and column $j$ of the second matrix. [I don't know what "vector dot product" is!] Finding $\text {H}_{1,1}$ $\text{H}_{1,1}$ is the dot product of the first row of $\text{E}$ and the first column of $\text{B}$. $ \text {H}=\left[\begin{array}{rr}{1} & {4} & {2} \\ 1 & 2 & 2\end{array}\right]\left[\begin{array}{rr} {1} & 3 \\ {2} & 3 \\ {2} & -2\end{array}\right]$ Therefore, this is the appropriate calculation of $\text{H}_{1,1}$. $\begin{aligned}\text{H}_{1,1}&=(1,4,2)\cdot(1,2,2)\\\\ &=1 \cdot 1+4\cdot 2 + 2\cdot 2\\\\ &=13 \end{aligned}$ The other entries of $\text{H}$ can be found similarly. Try it yourself for $\text{H}_{2,1}$ What is the appropriate calculation of ${H}_{2,1}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $4 \cdot 1+2\cdot 3 = 10$ (Choice B) B $1 \cdot 1+2\cdot 2 + 2\cdot 2 = 9$ (Choice C) C $1 \cdot 3+4\cdot 3 + 2\cdot -2 = 11$ Check Summary After calculating all the remaining entries of $\text{H}$, we get the following answer. $ \text {H}=\left[\begin{array}{rr}13 & 11 \\ 9 & 5\end{array}\right]$